Evaluate the following integrals:

Assume $\log (\tan x)=t$

$d(\log (\tan x))=d t$

$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=d t$

$\Rightarrow \sec x \cdot \operatorname{cosec} x \cdot d x=d t$

Put $\mathrm{t}$ and $\mathrm{dt}$ in given equation we get

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\ln |t|+c$

But $t=\log (\tan x)$

$=\ln |\log (\tan x)|+c$